RAINBOW - Editorial

PROBLEM LINK:

Practice
Contest

Author: Lewin Gan
Testers: Kamil Dębowski
Editorialist: Lewin Gan

DIFFICULTY:

Easy-Medium

PREREQUISITES:

Graphs

PROBLEM:

Find a maximum subset of nodes such that each node has at least two neighbors of different colors.

QUICK EXPLANATION:

Remove nodes that cannot be part of any interesting subset, and repeat.

EXPLANATION:

Let’s call a vertex “bad” if all its neighbors are the same color. Obviously, we can’t have any bad vertices in our interesting subset, so let’s remove them from the graph.
This might introduce some more bad vertices, but we can keep repeating this process (this is illustrated in the last sample case).

Now, if we have no bad vertices, we claim that the subset is interesting. Of course, this follows from the definition of an interesting subset. What about maximum size?
Well, we only removed vertices that had no way of being part of an interesting subset. Thus, we must have a subset of maximum size at the end since we never removed something that could be part of an interesting subset.

AUTHOR’S AND TESTER’S SOLUTIONS:

Setter
Tester

2 Likes

Tester should have used assert statements, to check whether the given constraints are satisfied.

1 Like

I did it. The thing is that we made some change in constraints one day before the contest and I didn’t modify my validator (program with asserts) — I will try not to let it happen again. I’m sorry for that.

4 Likes

I spend > 1hrs to debug and cannot get it accepted. Then suddenly, when I woke up in the next morning, my solution is ACed. Why?

There was a small error in the test files. It was fixed at the ending of the contest which was extended by 15 mins.

1 Like

Its okay dear, you did your best. :slight_smile:

the tester’s solution time complexity seems to be O(N^2) for the worst case .I wonder if there is some optimal approach for this, doesn’t seem to be existent .

i couldn’t understand the editorial properly problem says that color/a number is assigned to an edge.but editorial says that fiind neighbour node of different color!!! anyone please explain editorial

@forhadsidhu neigbour, since the graph is a complete graph .

@daljit1 actually i couldn’t understand the editorial.is it possiblem to explain what u underrstood??

The editorial is self explanatory :slight_smile:

We needed to find subset of veriticies so that all of them had edges from them not of same colour. If any vertex had all the edges from it of same colour to all vetricies in the subset, then such a vetrex is termed bad.So, keep removing bad vertices from the graph. Try solving the last test case given the examples.

The tester Solution is O(N^2)…can’t we do better than this…?

The setter’s solution is also O(N^2). Anyways, we can’t do better than this since we need at least O(N^2) time just to read the input.

1 Like

The setter’s solution is also O(N^2). We can’t do any better since we need O(N^2) time just to read the input.

2 Likes

Thankyou ,I was just wondering ;).

@forhadsidhu you start with the complete graph given to you. Each node has edges emanating off it. Let us denote the total number of distinctly colored edges that come out of node i as diff[i]. Now once you’ve performed this task remove all the nodes which have diff[i] < 2. However, as you are removing nodes it means other nodes might get affected and their values of diff[i] might decrease below 2. Suppose node i only has green colored edges coming out of it and node j has one green colored edge to node i and the rest are orange colored edges. Now when you calculate diff[i] you will find that node i needs to be removed but node j can stay. However, once you’ve removed node i node j will become faulty as now it has only orange colored edges coming off it. So you need to perform this process until you get a set of nodes where each node has diff[i] >= 2.
As for the proof of why this will work: Suppose a node has diff[i] < 2, it’s faulty. Then it can’t be a part of any set of nodes we give as the answer. Therefore removing it does no harm to our solution and we can safely remove it. As we started with the biggest set (the complete graph) the first time we encounter such a set of nodes where diff[i] >= 2, will be the biggest set with the required properties.

@forhadsidhu you start with the complete graph given to you. Each node has edges emanating off it. Let us denote the total number of distinctly colored edges that come out of node i as diff[i]. Now once you’ve performed this task remove all the nodes which have diff[i] < 2. However, as you are removing nodes it means other nodes might get affected and their values of diff[i] might decrease below 2. Suppose node i only has green colored edges coming out of it and node j has one green colored edge to node i and the rest are orange colored edges. Now when you calculate diff[i] you will find that node i needs to be removed but node j can stay. However, once you’ve removed node i node j will become faulty as now it has only orange colored edges coming off it. So you need to perform this process until you get a set of nodes where each node has diff[i] >= 2.
As for the proof of why this will work: Suppose a node has diff[i] < 2, it’s faulty. Then it can’t be a part of any set of nodes we give as the answer. Therefore removing it does no harm to our solution and we can safely remove it. As we started with the biggest set (the complete graph) the first time we encounter such a set of nodes where diff[i] >= 2, will be the biggest set with the required properties.

//